• # question_answer Let$z$and $w$ be two complex numbers such that $|z|\,\le 1,$ $|w|\,\le 1$and$|z+iw|\,=\,|z-i\overline{w}|=2$. Then $z$ is equal to [IIT 1995] A) 1 or $i$ B) $i$ or $-i$ C) 1 or - 1 D) $i$or -1

Let $z=a+ib,|z|\le 1$Þ ${{a}^{2}}+{{b}^{2}}\le 1$ and $w=c+id,|w|\,\le 1$Þ ${{c}^{2}}+{{d}^{2}}\le 1$      $|z+iw|\,=\,|a+ib+i(c+id)|=2$ Þ ${{(a-d)}^{2}}+{{(b+c)}^{2}}=4$ ......(i)      $|z-i\overline{w}|\,=\,|a+ib-i(c-id)|$ Þ ${{(a-d)}^{2}}+{{(b-c)}^{2}}=4$ ......(ii) From (i) and (ii), we get $bc=0$ Þ Either $b=0$or $c=0$ If$b=0$, then${{a}^{2}}\le 1$. Then, only possibility is $a=1$or$-1$.